这真的很简单。

if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;

(当然，为了简洁起见，我省略了一些程序。)如果程序员厌倦了把这些都输入，我相信他或她可以编写一个单独的程序来为他生成这些。我碰巧认识一个能大大简化他工作的人。

这是经典的2进制除法算法

#include <stdio.h>
#include <stdint.h>

int main()
{
  uint32_t mod3[6] = { 0,1,2,0,1,2 };
  uint32_t x = 1234567; // number to divide, and remainder at the end
  uint32_t y = 0; // result
  int bit = 31; // current bit
  printf("X=%u   X/3=%u\n",x,x/3); // the '/3' is for testing

  while (bit>0)
  {
    printf("BIT=%d  X=%u  Y=%u\n",bit,x,y);
    // decrement bit
    int h = 1; while (1) { bit ^= h; if ( bit&h ) h <<= 1; else break; }
    uint32_t r = x>>bit;  // current remainder in 0..5
    x ^= r<<bit;          // remove R bits from X
    if (r >= 3) y |= 1<<bit; // new output bit
    x |= mod3[r]<<bit;    // new remainder inserted in X
  }
  printf("Y=%u\n",y);
}

(注意:查看下面的编辑2以获得更好的版本!)

这并不像听起来那么棘手，因为你说“没有使用[..+[…]运营商”。如果你想禁止同时使用+字符，请参见下面。

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    for (unsigned i = 0; i < by; i++)
      cmp++; // that's not the + operator!
    floor = r;
    r++; // neither is this.
  }
  return floor;
}

然后用div_by(100,3)将100除以3。

编辑:你可以继续并替换++操作符:

unsigned inc(unsigned x) {
  for (unsigned mask = 1; mask; mask <<= 1) {
    if (mask & x)
      x &= ~mask;
    else
      return x & mask;
  }
  return 0; // overflow (note that both x and mask are 0 here)
}

编辑2:稍快的版本，不使用任何包含+、-、*、/、%字符的操作符。

unsigned add(char const zero[], unsigned const x, unsigned const y) {
  // this exploits that &foo[bar] == foo+bar if foo is of type char*
  return (int)(uintptr_t)(&((&zero[x])[y]));
}

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    cmp = add(0,cmp,by);
    floor = r;
    r = add(0,r,1);
  }
  return floor;
}

我们使用add函数的第一个参数，因为不使用*字符就不能表示指针的类型，除非在函数形参列表中，其中的语法类型[]与类型* const相同。

FWIW，你可以很容易地实现一个乘法函数使用类似的技巧使用0x55555556技巧提出的AndreyT:

int mul(int const x, int const y) {
  return sizeof(struct {
    char const ignore[y];
  }[x]);
}

通过使用eval和字符串连接来在“幕后”使用/操作符是欺骗吗?

例如，在Javacript中，您可以这样做

function div3 (n) {
    var div = String.fromCharCode(47);
    return eval([n, div, 3].join(""));
}

使用fma()库函数的解决方案，适用于任何正数:

#include <stdio.h>
#include <math.h>

int main()
{
    int number = 8;//Any +ve no.
    int temp = 3, result = 0;
    while(temp <= number){
        temp = fma(temp, 1, 3); //fma(a, b, c) is a library function and returns (a*b) + c.
        result = fma(result, 1, 1);
    } 
    printf("\n\n%d divided by 3 = %d\n", number, result);
}

请看我的另一个答案。

这是Python中的，基本上，字符串比较和一个状态机。

def divide_by_3(input):
  to_do = {}
  enque_index = 0
  zero_to_9 = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
  leave_over = 0
  for left_over in (0, 1, 2):
    for digit in zero_to_9:
      # left_over, digit => enque, leave_over
      to_do[(left_over, digit)] = (zero_to_9[enque_index], leave_over)
      if leave_over == 0:
        leave_over = 1
      elif leave_over == 1:
        leave_over = 2
      elif leave_over == 2 and enque_index != 9:
        leave_over = 0
        enque_index = (1, 2, 3, 4, 5, 6, 7, 8, 9)[enque_index]
  answer_q = []
  left_over = 0
  digits = list(str(input))
  if digits[0] == "-":
    answer_q.append("-")
  digits = digits[1:]
  for digit in digits:
    enque, left_over = to_do[(left_over, int(digit))]
    if enque or len(answer_q):
      answer_q.append(enque)
  answer = 0
  if len(answer_q):
    answer = int("".join([str(a) for a in answer_q]))
  return answer